simply supported beam

SIMPLY SUPPORTED FLANGED BEAM
DESIGN SIMPLY SUPPORTED FLANGED BEAM

bf

1) Load Analysis
- N= 1.35gk + 1.5qk
2) SFD and BMD
- consider type of load

hf

h

*min diameter bar provided is 12mm
*min diameter link provided is 8mm

d

d = h – Cnom – Ølink – Øbar/2

Neutral Axis Lies in Flange
Design as a rectangular section
Size of beam (bf X d)

Z = d (0.5+(0.25 – (k/1.134))1/2 0.95 d, use 0.95d as z value

Asreq = M/0.87fykZ
Provide main reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

Neutral Axis Below Flange
Design as a flange section

Muf = 0.167fckbwd2+ 0.567 fck(bf–bw) (d-(hf/2))
If M < Muf – singly reinforced
If M > Muf – doubly reinforced

K = M/fckbd2
If k < 0.167 (singly reinforced)
If k > 0.167 (doubly reinforced)
SINGLY

d’ = Cnom + Ølink + Øbar/2

Mf = (0.567fckbfhf)(d-(hf/2))
M < Mf – neutral axis lies in flange
M > Mf – neutral axis below flange

bw

SINGLY

DOUBLY

Compression reinforcement:
As’ = (K-Kbal)fckbd2/0.87fyk(d-d’) d’ = Cnom + Ølink – Øbar
Provide compression reinforcement

Tension reinforcement:
As = (Kbalfckbd2/0.87fykZ) + As’
Provide tension reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

DOUBLY

Asreq= M + 0.1 fckbwd (0.36d –hf)/0.87fyk(d-(hf/2))
Provide main tension reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

As’ = (M-Muf)/0.87 fyk(d-d’)
Provide main compression reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

As = [0.2 fckbwd + 0.567 fckhf(bf-bw) ] + As’
0.87 fyk
Provide main tension reinforcement
Asmin = 0.26fctmbd/fyk < 0.0013bd
Asmax = 0.04Ac

*IF FORMULA IN APPENDIX IS DIFFERENT, YOU MAY FOLLOW APPENDIX.

MSAH/FKA/UiTMPP

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SIMPLY SUPPORTED FLANGED BEAM
SHEAR REINFORCEMENT DESIGN:
VEd = Vmax – from SFD
VRd,c = 0.12k(100ρ1fck)1/3bwd k = 1 + (200/d)1/2 < 2.0mm if more than 2.0 used 2.0 as k ρ1 = Asl/bwd Asl = Area of tension reinforcement provided at mid-span
Vmin